What is the concentration of a #KOH (aq)# solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L #H_2SO_4(aq)#?

1 Answer
Jan 17, 2016

Answer:

We assume stoichiometric quantities of potassium hydroxide and sulfuric acid.

Explanation:

#2KOH(aq) + H_2SO_4 rarr K_2SO_4(aq) + 2H_2O(l)#

The reaction represents the neutralization of sulfuric acid by potassium hydroxide.

Moles of sulfuric acid: #25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)# #=# #??# #mol#

From the equation above, which shows that the stoichiometry of hydroxide to sulfuric acid is #2:1#, and the volume of stoichiometric acid:

#[KOH]=(25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)xx2)/(12.8xx10^-3L)# #=# #??mol*L^-1#.

We get an answer in #mol*L^-1#, as required.